Question

Find solution of the system of differential equations in matrix form using eigenvalues and eigenvectors: `x'=([3,1],[1,3])x`.

Answer

`x'=([3,1],[1,3])x`.

First, find eigenvalues of the matrix `A=([3,1],[1,3])`:

`det(A-rI)=0` where `rI=([r,0],[0,r])`.

`|(3-r,1),(1,3-r)|=0`.

`(3-r)(3-r)-1*1=0`.

`r^2-6r+9-1=0`.

`r^2-6r+8=0`.

`(r-2)(r-4)=0`.

`r_1=2` and `r_2=4`.

Now, for each eigenvalue find corresponding eigenvector:

  1. `r_1=2`.

    `(A-r_1 I)v_1=0`.

    `([3-2,1],[1,3-2])v_1=0`.

    `([1,1],[1,1])v_1=0`.

    `([1,1],[0,0])v_1=0`.

    So, `v_1=((k),(-k))`. If we take `k=1` then `v_1=((1),(-1))`.

  2. `r_1=4`.

    `(A-r_2 I)v_2=0`.

    `([3-4,1],[1,3-4])v_2=0`.

    `([-1,1],[1,-1])v_2=0`.

    `([-1,1],[0,0])v_2=0`.

    So, `v_2=((c),(c))`. If we take `c=1` then `v_2=((1),(1))`.

So,

`x=c_1e^(2t)([1],[-1])+c_2e^(4t)([1],[1])=([c_1e^(2t)+c_2e^(4t)],[-c_1e^(2t)+c_2e^(4t)])`.

Answer: `x=([c_1e^(2t)+c_2e^(4t)],[-c_1e^(2t)+c_2e^(4t)])`.